Question

In a recent nationwide survey of 1317 persons 18 and over in the United States, 596 of the respondents reported that acquiring material wealth was "of significant importance in their lives. Calculate a 95% confidence interval for the true proportion of adults in the United States who believe acquiring wealth is 'of significant importance to three decimal places. Take all calculations toward the answer to four (4) decimal places, and choose the closest answer 0.418 to 0.488 0.439 to 0.466 ООО 0.426 to 0.480 0.430 to 0.488

Answer 1

Solution:
Given in the question
Number of sample (n) = 1317
Number of favourable cases(X) = 596
Sample proportion(p^) or point estimate = X/n = 596/1317 = 0.4525
We need to calculate 95% confidence interval can be calculated as
p^ +/- Z$\alpha$/2 * sqrt(p^ * (1-p^)/n)
Confidence level = 0.95
level of significance($\alpha$) = 0.05, $\alpha$ /2 = 0.025
From Z table we found Z$\alpha$/2 = 1.96
So 95% confidence interval is
0.4525 +/- 1.96*sqrt(0.4525*(1-0.4525)/1317)
0.4525 +/- 1.96*0.0137
0.4525 +/- 0.027
So 95% confidence interval is 0.426 to 0.480
So its correct answer is C i.e. 0.426 to 0.480